A gas mixture containing 0.20\text{ mol} of oxygen (\text{O}_2) and 0.30\text{ mol} of nit… — JAMB UTME

A gas mixture containing $0.20\text{ mol}$ of oxygen ( $\text{O}_2$ ) and $0.30\text{ mol}$ of nitrogen ( $\text{N}_2$ ) is collected over water at $27^\circ\text{C}$ . The total pressure of the moist gas mixture is $775\text{ mmHg}$ and the saturated vapor pressure of water at $27^\circ\text{C}$ is $25\text{ mmHg}$ . If $0.50\text{ mol}$ of dry argon ( $\text{Ar}$ ) gas is subsequently added to the mixture at the same temperature while maintaining the total system pressure at $775\text{ mmHg}$ , what is the final partial pressure of nitrogen gas in the moist mixture in $\text{mmHg}$ ?

Answer: 225 mmHg

Answer

The final partial pressure of nitrogen gas in the moist mixture is

225\text{ mmHg}

When a gas is collected over water, the total pressure measured includes the saturated vapor pressure of water (

P_{\text{H}_2\text{O}}

). Subtracting

P_{\text{H}_2\text{O}} = 25\text{ mmHg}

from

P_{\text{total}} = 775\text{ mmHg}

yields the total pressure of the dry gases (

P_{\text{dry}} = 750\text{ mmHg}

). Adding

0.50\text{ mol}

of argon increases the total dry gas quantity to

1.00\text{ mol}

(

0.20 + 0.30 + 0.50

). The mole fraction of nitrogen in this dry mixture is

\frac{0.30}{1.00} = 0.30

. Finally, multiplying this mole fraction by

P_{\text{dry}}

gives the partial pressure of nitrogen as

0.30 \times 750\text{ mmHg} = 225\text{ mmHg}

Step-by-Step Solution

Calculate the total pressure exerted by the dry gas mixture.

P_{\text{dry}} = 750\text{ mmHg}

According to Dalton's Law of Partial Pressures, the total pressure of a gas collected over water is

P_{\text{total}} = P_{\text{dry}} + P_{\text{H}_2\text{O}}

. Therefore,

P_{\text{dry}} = 775\text{ mmHg} - 25\text{ mmHg} = 750\text{ mmHg}

Calculate total moles of dry gas after argon addition.

n_{\text{total, dry}} = 1.00\text{ mol}

Sum the moles of oxygen (

0.20\text{ mol}

), nitrogen (

0.30\text{ mol}

), and added argon (

0.50\text{ mol}

0.20 + 0.30 + 0.50 = 1.00\text{ mol}

Determine the mole fraction of nitrogen gas in the dry mixture.

\chi_{\text{N}_2} = 0.30

The mole fraction is the ratio of the moles of nitrogen to the total moles of dry gas:

\frac{0.30\text{ mol}}{1.00\text{ mol}} = 0.30

Calculate the partial pressure of nitrogen gas.

P_{\text{N}_2} = 225\text{ mmHg}

The partial pressure of an individual gas component in a dry mixture is given by

P_{\text{N}_2} = \chi_{\text{N}_2} \times P_{\text{dry}} = 0.30 \times 750\text{ mmHg} = 225\text{ mmHg}

Key Concept

Dalton's Law of Partial Pressures with Vapor Pressure Correction and Mole Fraction Calculation

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